Question

Solve the following equation for B, where -A

Answer 1

Given:

`5\tan ^2\beta+5\tan \beta=0`
`5\tan \beta(\tan \beta+1)=0`
`\begin{gathered} 5\tan \beta=0 \\ \tan \beta=0 \\ \beta=n\pi \end{gathered}`
`\begin{gathered} \tan \beta+1=0 \\ \tan \beta=-1 \\ \beta=(3\pi)/(4)+n\pi \end{gathered}`
`\beta=n\pi,(3\pi)/(4)+n\pi\text{ , where n is any integer}`

When n=0,

`\beta=0,(3\pi)/(4)`

When n= -1

`\beta=-\pi,-(\pi)/(4)`

When n= 1

`\beta=\pi,(7\pi)/(4)\text{ }\Rightarrow\text{ Not valid because }-\pi\le\beta<\pi`

Therefore the final answer is

`\beta=-\pi,-(\pi)/(4),0,(3\pi)/(4)`